Since the Borel-\(\sigma \)  algebra \(\mathcal {B}(\mathbb {R}^2)\)  is equal to \(\mathcal {B}(\mathbb {R})\otimes \mathcal {B}(\mathbb {R})\)  it
suffices to show that \(M\)  and \(a\)  are Borel-measurable. By right-continuity of \(f \in D[0,1]\)  we have that \(M(f)=\sup _{t \in [0,1] \cap \mathbb {Q}}f(t)\)  and
therefore \(M:(D[0,1],s)\rightarrow \mathbb {R}\)  is Borel-measurable upon noticing that the Borel-\(\sigma \)  algebra on \((D[0,1],s)\)  is generated by the
projections (evaluation maps), see Theorem 12.5 in [1]}.
